LECTURE NOTES ON TRIPLE PRODUCT PROF. SHILPA PATRA Department of Mathematics Narajole Raj College 1. Triple Product In vector algebra, a branch of mathematics, the triple product is a product of three 3-dimensional vectors, usually Euclidean vectors. The name "triple product" is used for two different products, the scalar-valued scalar triple product and the vector-valued vector triple product. 2. Scalar triple product The scalar triple product (also called the mixed product, box product, or triple scalar product) is defined as the dot product of one of the vectors with the cross product of the other two. Definition 2.1. For a given set of three vectors a; b; and c, the scalar (a×b)·c is called a scalar triple product of a; b; and c. Remark: a · b is a scalar, and so (a · b) × c has no meaning. Note: Given any three vectors a; b; and c, the following are scalar triple prod- ucts: (a × b) · c; (b × c) · a; (c × a) · b; a · (b × c); b · (c × a); c · (a × b) (2.1) (b × a) · c; (c × b) · a; (a × c) · b; a · (c × b); b · (a × c); c · (b × a) Geometric interpretation of scalar triple product: 1 Paper:C4T Unit-4 Topic- Triple product Geometrically, the scalar triple product (a × b) · c is the (signed) volume of the parallelepiped defined by the three vectors given. Indeed, the magnitude of the vector a × b is the area of the parallelogram formed by using a and b; and the direction of the vector a × b is perpendicular to the plane parallel to both a and b. Therefore, j(a × b) · cj is ja × bjjcjj cos(θ)j, where θ is the angle between a × b and c. From the above figure, we observe that jcjj cos(θ)j is the height of the paral- lelepiped formed by using the three vectors as adjacent vectors. Thus, j(a × b) · cj is the volume of the parallelepiped. Theorem 2.2. If a = a1i+a2j+a3k, b = b1i+b2j+b3k, and c = c1i+c2j+c3k, then a1 a2 a3 (a × b) · c = b1 b2 b3 : c1 c2 c3 Paper:C4T Unit-4 Topic- Triple product Proof. By definition, we have i j k (a × b) · c = a1 a2 a3 · c b1 b2 b3 = [(a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k] · (c1i + c2j + c3k) = (a2b3 − a3b2)c1 + (a3b1 − a1b3)c2 + (a1b2 − a2b1)c3 a1 a2 a3 = b1 b2 b3 : c1 c2 c3 Theorem 2.3. For a given set of three vectors a; b; and c, (a × b) · c = a · (b × c): Proof. b1 b2 b3 a · (b × c) = (b × c) · a = c1 c2 c3 a1 a2 a3 a1 a2 a3 = − c1 c2 c3 by R1 $ R3 b1 b2 b3 a1 a2 a3 = b1 b2 b3 by R2 $ R3 c1 c2 c3 = (a × b) · c Note: By Theorem 2.3, it follows that, in a scalar triple product, dot and cross can be interchanged without altering the order of occurrences of the vectors. For instance, we have (a × b) · c = a · (b × c); Since dot and cross can be interchanged: = (b × c) · a; Since dot product is commutative: = b · (c × a); Since dot and cross can be interchanged: = (c × a) · b; Since dot product is commutative: = c · (a × b); Since dot and cross can be interchanged: Notation: For any three vectors a; b; and c, the scalar triple product (a × b) · c is denoted by [a; b; c] is read as box a, b, c . For this reason and also because the absolute value of a scalar triple product represents the volume of a box (rectangular parallelepiped),a scalar triple product is also called a box product. Note: By the above, we have [a; b; c] = (a × b) · c = a · (b × c) = (b × c) · a = b · (c × a) = [b; c; a] [b; c; a] = (b × c) · a = b · (c × a) = (c × a) · b = c · (a × b = [c; a; b]: Paper:C4T Unit-4 Topic- Triple product In other words, [a; b; c] = [b; c; a] = [c; a; b] ; that is, if the three vectors are per- muted in the same cyclic order, the value of the scalar triple product remains the same. If any two vectors are interchanged in their position in a scalar triple product, then the value of the scalar triple product is (-1) times the original value. More explicitly, [a; b; c] = [b; c; a] = [c; a; b] = −[a; c; b] = −[c; b; a] = −[b; a; c]: Theorem 2.4. The scalar triple product preserves addition and scalar multipli- cation. That is, [a + b; c; d] = [a; c; d] + [b; c; d]; [αa; b; c] = α[a; b; c]; 8α 2 R [a; b + c; d] = [a; b; d] + [a; c; d]; [a; αb; c] = α[a; b; c]; 8α 2 R [a; b; c + d] = [a; b; c] + [a; b; d]; [a; b; αc] = α[a; b; c]; 8α 2 R Theorem 2.5. The scalar triple product of three non-zero vectors is zero if, and only if, the three vectors are coplanar. Proof. Leta; b; c be any three non-zero vectors. Then, (a × b) · c = 0 () a × b = 0 or (c and a × b are perpendicular) () (a and b are parallel) or (a; b; and c are coplanar) () a; b; and c are coplanar: Example 2.6. If 2i − j + 3k, 3i + 2j + k, i + mj + 4k are coplanar, find the value of m . Proof. Since the given three vectors are coplanar, we have 2 −1 3 3 2 1 = 0 =) m = −3 1 m 4 Example 2.7. Show that the four points (6, 7, 0), (16, 19, 4), (0, 3, 6), (2, 5,10) lie on a same plane. Proof. Let A = (6, 7, 0), B = (16, 19, 4), C = (0, 3, 6), D = (2, 5,10) . To show that the four points A, B, C, D lie on a plane, we have to prove that the three −−! −! −−! vectors AB, AC and AD are coplanar. Now, −−! −−! −! AB = OB − OA = (16i − 19j − 4k) − (6i − 7j) = (10i − 12j − 4k) −! −−! −! −−! −−! −! AC = OC − OA = (−6i + 10j − 6k) and AD = OD − OA = (−4i + 2j + 10k). we have 10 −12 −4 −−! −! −−! [AB; AC; AD] = −6 10 −6 = 0: −4 2 10 Paper:C4T Unit-4 Topic- Triple product −−! −! −−! Therefore, the three vectors AB, AC and AD are coplanar and hence the four points A, B, C, and D lie on a plane. Example 2.8. If the vectors a; b; c are coplanar, then prove that the vectors a + b; b + c; c + a are also coplanar. Proof. Since the vectors a; b; c are coplanar, we have [a; b; c] = 0 Using the prop- erties of the scalar triple product, we get [a + b; b + c; c + a] = [a; b + c; c + a] + [b; b + c; c + a] = [a; b; c + a] + [a; c; c + a] + [b; b; c + a] + [b; c; c + a] = [a; b; c] + [a; b; a] + [a; c; c] + [a; c; a] + [b; b; c] + [b; b; a] + [b; c; c] + [b; c; a] = [a; b; c] + [a; b; c] = 2[a; b; c] = 0: Hence the vectors a + b; b + c; c + a are coplanar. Theorem 2.9. If a; b; c and p; q; r be any systems of three vectors, and if p = x1a + y1b + z1c, q = x2a + y2b + z2c, and r = x3a + y3b + z3c, then x1 y1 z1 [p; q; r] = x2 y2 z2 [a; b; c]: x3 y3 z3 Proof. Applying the distributive law of cross product and using a × a = b × b = c × c = 0 b × a = −a × b; a × c = −c × a; c × b = −b × c: We get p × q = (x1a + y1b + z1c) × (x2a + y2b + z2c) = (x1y2 − x2y1)(a × b) + (y1z2 − y2z1)(b × c) + (z1x2 − z2x1)(c × a) x1 x2 y1 y2 z1 z2 = (a × b) + (b × c) + (c × a): y1 y2 z1 z2 x1 x2 Hence, we get [p; q; r] = (p × q) · (x3a + y3b + z3c) y1 y2 z1 z2 x1 x2 = x3 + y3 + z3 [a; b; c] z1 z2 x1 x2 y1 y2 y1 z1 z1 x1 x1 y1 = x3 + y3 + z3 [a; b; c] y2 z2 z2 x2 x2 y2 x1 y1 z1 = x2 y2 z2 [a; b; c]: x3 y3 z3 Example 2.10. If a; b; c are three vectors, prove that [a + c; a + b; a + b + c] = −[a; b; c]: Paper:C4T Unit-4 Topic- Triple product Proof. Using theorem 2.9, we get 1 0 1 [a + c; a + b; a + b + c] = 1 1 0 [a; b; c] 1 1 1 = −[a; b; c]: Properties(Summary): • The scalar triple product is unchanged under a circular shift of its three operands (a, b, c): a · (b × c) = b · (c × a) = c · (a × b): • Swapping the positions of the operators without re-ordering the operands leaves the triple product unchanged. This follows from the preceding prop- erty and the commutative property of the dot product. a · (b × c) = (a × b) · c: • Swapping any two of the three operands negates the triple product. This follows from the circular-shift property and the anticommutativity of the cross product. a · (b × c) = −a · (c × b) = −b · (a × c) = −c · (b × a): • The scalar triple product can also be understood as the determinant of the 3 × 3 matrix that has the three vectors either as its rows or its columns 2 3 a1 a2 a3 a · (b × c) = det 4b1 b2 b3 5 = det (a; b; c) : c1 c2 c3 • If the scalar triple product is equal to zero, then the three vectors a, b, and c are coplanar, since the parallelepiped defined by them would be flat and have no volume.